Let's say I have a following:
>>tmp='1 1 1 1 1 -1 -1 -1 -1 -1'
>>echo $tmp
1 1 1 1 1 -1 -1 -1 -1 -1
And I use the commands:
>>echo $tmp | grep -ow 1 | wc -l
10
>>echo $tmp | grep -ow "\-1" | wc -l
5
How am I able to get just the counts of 1 (which the answer should be 5 given the example above) without including the negative 1's?
You may use
echo "$temp" | grep -oE '(^|[^-0-9])1\b' | wc -l
Or, if the numbers are separated with whitespace, use whitespace boundaries you may use a PCRE regex with a GNU grep, or a Perl equivalent:
echo "$temp" | grep -oP '(?<!\S)1(?!\S)' | wc -l
perl -lne 'END {print $c} map ++$c, /(?<!\S)1(?!\S)/g'
See the online demo #1 and online demo #2.
Details
-o - output matches only
-E - enable POSIX ERE syntax
-P - enables PCRE syntax
(^|[^-0-9]) - matches start of string (^) or (|) a char other than - and a digit
(?<!\S) - left-hand whitespace boundary
1 - a 1 digit
\b - a word boundary
(?!\S) - right-hand whitespace boundary
Related
Problem
I want to get any text that consists of 1 to three digits followed by a % but without the % using sed.
What I tried
So i guess the following regex should match the right pattern : [0-9]{1,3}%.
Then i can use this sed command to catch the three digits and only print them :
sed -nE 's/.*([0-9]{1,3})%.*/\1/p'
Example
However when i run it, it shows :
$ echo "100%" | sed -nE 's/.*([0-9]{1,3})%.*/\1/p'
0
instead of
100
Obviously, there's something wrong with my sed command and i think the problem comes from here :
[0-9]{1,3}
which apparently doesn't do what i want it to do.
edit:
Solution
The .* at the start of sed -nE 's/.*([0-9]{1,3})%.*/\1/p' "ate" the two first digits.
The right way to write it, according to Wicktor's answer, is :
sed -nE 's/(.*[^0-9])?([0-9]{1,3})%.*/\2/p'
The .* grabs all digits leaving just the last of the three digits in 100%.
Use
sed -nE 's/(.*[^0-9])?([0-9]{1,3})%.*/\2/p'
Details
(.*[^0-9])? - (Group 1) an optional sequence of any 0 or more chars up to the non-digit char including it
([0-9]{1,3}) - (Group 2) one to three digits
% - a % char
.* - the rest of the string.
The match is replaced with Group 2 contents, and that is the only value printed since n suppresses the default line output.
It will be easier to use a cut + grep option:
echo "abc 100%" | cut -d% -f1 | grep -oE '[0-9]{1,3}'
100
echo "100%" | cut -d% -f1 | grep -oE '[0-9]{1,3}'
100
Or else you may use this awk:
echo "100%" | awk 'match($0, /[0-9]{1,3}%/){print substr($0, RSTART, RLENGTH-1)}'
100
Or else if you have gnu grep then use -P (PCRE) option:
echo "abc 100%" | ggrep -oP '[0-9]{1,3}(?=%)'
100
This might work for you (GNU sed):
sed -En 's/.*\<([0-9]{1,3})%.*/\1/p' file
This is a filtering exercise, so use the -n option.
Use a back reference to capture 1 to 3 digits, followed by % and print the result if successful.
N.B. The \< ensures the digits start on a word boundary, \b could also be used. The -E option is employed to reduce the number of back slashes which would normally be necessary to quote (,),{ and } metacharacters.
I have the following text:
(took 1.22 seconds)
need to replace the time to TIME using regex.
I want the string to be:
(took TIME seconds)
how can I do it?
I am using a script in Unix similar to the following:
's/user detected .*$/user detected USER/g'
You may use any of the sed solutions below:
s='took 1.22 seconds'
echo $s | sed 's/ [0-9]*\.\?[0-9]\{2\} / TIME /g'
echo $s | sed -E 's/[0-9]*\.?[0-9]{2}/TIME/g'
echo $s | sed -E 's/\b[0-9]*\.?[0-9]{2}\b/TIME/g'
See the online demo
The patterns mean:
[0-9]* - zero or more digits
\.? - 1 or 0 dots
[0-9]{2} - 2 digits
\b - word boundary.
I would like to extract 1, 10, and 100 from:
1 one -args 123
10 ten -args 123
100 one hundred -args 123
However this regex returns 100:
echo -e " 1 one\n 10 ten\n100 one hundred" | grep -Po '^(?=[ ]*)\d+(?=.*)'
100
Not ignoring the preceding spaces returns the numbers (but of course with undesired spaces):
echo -e " 1 one\n 10 ten\n100 one hundred" | grep -Po '^[ ]*\d+(?=.*)'
1
10
100
Have I misunderstood non capturing regex groups in grep / Perl (grep version 2.2, Perl as the -P flag should use its regex) or is this a bug? I notice the release notes for 2.6 says "This release fixes an unexpectedly large number of flaws, from outright bugs (surprisingly many, considering this is "grep")".
If someone with 2.6 could try these examples that would be valuable to determine if this is a bug (in 2.2) or intended behaviour.
The issue is what is considered a 'match' by grep. In the absence of telling grep part of the total match is not what you want, it prints everything up to the end of the match regardless of matching groups.
Given:
$ echo "$txt"
1 one -args 123
10 ten -args 123
100 one hundred -args 123
You can get just the first column of digits without leading spaces several ways.
With GNU grep:
$ echo "$txt" | grep -Po '^[ ]*\K\d+'
1
10
100
Here \K is equivalent to a look behind assertion that resets the match text of the match to be what comes after. The left hand, before the \K, is required to match, but is not included in match text printed by grep.
Demo
awk:
$ echo "$txt" | awk '/^[ ]*[0-9]+/{print $1}'
sed:
$ echo "$txt" | sed 's/^[ ]*\([0-9]*\).*/\1/'
Perl:
$ echo "$txt" | perl -lne 'print $1 if /^[ ]*\K(\d+)/'
And then if you want the matches on a single line, run through xargs:
$ echo "$txt" | grep -Po '^[ ]*\K(\d+)' | xargs
1 10 100
Or, if you are using awk or Perl, just change the way it is printed to not include a carriage return.
You can delete the unwanted spaces this way :
echo -e " 1 one\n 10 ten\n100 one hundred" | grep -Po '^[ ]*(\d+)' | tr -d ' '
As for your question of why it is not working, it is not a bug, it is working as intended, you just misinterpreted how it should work.
If we focus on this ^(?=[ ]*)\d+:
The (?=[ ]*) part is a lookahead assertion. So it means that the regex engine tries to check if the ^ is followed by zero or more spaces. But the assertion itself is not part of the match, so in reality this code means :
- Match a ^ that is followed by 0 or more spaces
- After this ^, match one or more digits
So your code will only match when a digit is the first character of the line. The lookahead won't help you on your use case.
I think the anchor messes with the lookahead, which could be a lookbehind, but they can't be ambiguous (I always run into that one). So the following would work:
echo -e " 1 one\n 10 ten\n100 one hundred" | grep -Po '(?=[ ]*)\d+(?=.*)'
As for a better tool, I would use awk as it is suited to any column driven data. So if you were running it off of ps you could do something like:
ps | awk '/stuff you want to look for here/{print $1}'
awk will take care of all the white space by default
How should I match the first two bits (first occurrence only) in a digital stream line (1 byte line) using grep, in one direction only, i.e. (01 but not 10 in 01051015);
I've been tested:
grep -E '^[0-9]\{2\}$' | grep -Po --color '01' <<< 01051015
> 010-10-- (current output)
$cat -n test.txt
1 0001021113
2 0202031011
3 0103031113
4 ..........
$ grep -oE '^[0-1][0-9]\{2,2\}' | grep -E '(10)' ./test.txt > matchedList.txt
$ cat -n matchedList.txt
1 0001021113
2 0202031011
3 0103031113
But I need to parse and math the first "par occurrence", (in this case '10') ... in that specific order and one direction (like in line 2); so the correct utput should be: 2 0202031011
Tkx in advance
L.
grep -m 1 -e '^01' YourFile
where:
01 is your first 2 bit to find
-m 1 limit to 1st occurence
I have a list:
/device1/element1/CmdDiscovery
/device1/element1/CmdReaction
/device1/element1/Direction
/device1/element1/MS-E2E003-COM14/Field2
/device1/element1/MS-E2E003-COM14/Field3
/device1/element1/MS-E2E003-COM14/NRepeatLeft
How can I grep so that the returned strings containing only "Field" followed by digits or simply NRepeatLeft at the end of string (in my example it will be the last three strings)?
Expected output:
/device1/element1/MS-E2E003-COM14/Field2
/device1/element1/MS-E2E003-COM14/Field3
/device1/element1/MS-E2E003-COM14/NRepeatLeft
Try doing this :
grep -E "(Field[0-9]*|NRepeatLeft$)" file.txt
| | | ||
| | OR end_line |
| opening_choice closing_choice
extented_grep
if you don't have -E switch (stands for ERE : Extented Regex Expression):
grep "\(Field[0-9]*\|NRepeatLeft$\)" file.txt
OUTPUT
/device1/element1/MS-E2E003-COM14/Field2
/device1/element1/MS-E2E003-COM14/Field3
/device1/element1/MS-E2E003-COM14/NRepeatLeft
That will grep for lines matching Field[0-9] or lines matching RepeatLeft at the end. Is it what you expect ?
I am not much sure of how to use grep for your purpose.Probably you would like perl for this:
perl -lne 'if(/Field[\d]+/ or /NRepeatLeft/){print}' your_file
$ grep -E '(Field[0-9]*|NRepeatLeft)$' file.txt
Output:
/device1/element1/MS-E2E003-COM14/Field2
/device1/element1/MS-E2E003-COM14/Field3
/device1/element1/MS-E2E003-COM14/NRepeatLeft
Explanation:
Field # Match the literal word
[0-9]* # Followed by any number of digits
| # Or
NRepeatLeft # Match the literal word
$ # Match the end of the string
You can see how this works with your example here.